What price should the promotion be set at to generate the maximum daily profit?

Question

A local restaurant recently began to offer an eat-in pizza promotion for their famous double-decker pizza. When priced at $17.99/person, the promotion attracted 112 customers per day on average. When they raised the price to $19.99/person they attracted an average of only 96 customers per day.

What price should the promotion be set at to generate the maximum daily profit?

Please include calculations that include quadratic functions in your suggestion, Thank you!

Answer 1

$2 increase causes loss of 16 customers

$17.99 to $19.99 is a price rise of $2

112 -96 = 16 lost customers

a linear demand function would be a line through the points (112, 17.99) and (96, 19.99)

the line through them has slope = "rise over run" = (19.99-17.99)/(96-112) = 2/-16 = - 1/8

the equation is y = -(x-112)/8 + 17.99

where y = price and x = customers

y intercept = 255.92

x intercept = 31.99

divide by 2

(255.92/2, 31.99/2) = 127.96, 15.995)

= about (128,16)

Price of $16 has 128 customers

Maximum Revenue = 16(128) = about $2048

using more accurate numbers, not rounded off gives

Price of $15.995 gives max revenue = $2046.72= closer to $2047

that's max profit, if costs are zero

otherwise,

it's impossible to calculate max profit if costs are unknown

or using differential calculus with a quadratic revenue function

y = -(x-112)/8 + 17.99

yx = revenue = R = -(x^2-112x)/8 +17.99x

set R'(x) = 0 and solve for x

R = -x^2/8 +14x +17.99x= -x^2/8 +31.99x

R' = -x/4 + 31.99 = 0

x = 4(31.99) = 127.96 = about 128 customers

y = -(127.96 - 112)/8 + 17.99

= -(15.96)/8 +17.99

= 17.99-1.995

= 15.995

= about $16 = revenue maximizing price

= -8.245 +17.99

= 9.756

= about $9.76